Right? (I may seem like I'm answering my own question, but these Cambridge STEP questions do not give answers, so I'm basically working it out myself too)
Don't worry - it turns out that the problem is quite hard using GCSE maths. AS and A-level is nothing like this - they'll as good as tell you exactly what to do in each question I'll give you a step-by-step answer instead - do you know about partial differentiation? (Or any differentiation at all - we can work with that.)
I did add maths, so I know basic differentiation (x^2 goes to 2x, x goes to 1, 2 disappears). Never heard of partial differentiation
Silly question but why does the LHS become that by the chain rule?
I got this dxd(e)logx=logx(e)logx−1.0 ? which equals 0 haha
You're not differentiating correctly. dxdf(g(x))=g′(x)f′(g(x)); substitute g(x)=log(x),f(x)=exp(x) and solve for g'(x). What you have worked out is dedexp(log(x)), which is odd indeed given that e is constant.
Right? (I may seem like I'm answering my own question, but these Cambridge STEP questions do not give answers, so I'm basically working it out myself too)
That is incorrect. Can you see why this is the case? Using a quick check we can see that A would never have any reason to choose the bottom row (clearly something is wrong as the question asserts it must be reduced to a 2x2 matrix!) We also notice that there we can place a straight line that separates negative and positive terms.
Your error was in your 3rd dominance argument.
The desired 2x2 matrix is:
4 -5 -2 3
This clearly cannot be reduced by using dominance arguments. The most intuitive way to see this is to think of it as an array of arrows based on what numbers are higher than others.
In this case, we can visualize there as being four arrows whereby each pair points to opposite corners. Such as array necessary for irreducibility as there are four arrows and only four possible directions with which to point. A sufficient condition would also include as array whereby the 'arrows' all point in to the tail of the proceeding one. Alternatively, you could compare 'minimax' and 'maximin' to establish 'Nash equilibrium'.
We can then go on to solve this as a linear programming problem for A by letting the proportion of the time it plays each row be p and 1-p respectively. We could also use simplex if we wanted (though, in this case, this seems rather unnecessary).
I won't write the solution up. If you want to I can check it for you?
You're not differentiating correctly. dxdf(g(x))=g′(x)f′(g(x)); substitute g(x)=log(x),f(x)=exp(x) and solve for g'(x). What you have worked out is dedexp(log(x)), which is odd indeed given that e is constant.
The question is: Show that the identity is unique in a group. That is, show that if there are two elements e,f such that e+a = a+e = f+a = a+f = a for all a in the set X, then e=f.
This may take you a while to get your head around, but it's a good exercise in exploring a definition. That skill is very much not taught at school - even at A-level - so don't be disheartened if you don't know what to do. Write things down and see if they help! Good luck!
I see your valiant attempt at introducing us to groups has been ignored.
For your question at the end is my answer satisfactory?
e + a = f + a e + a + z = f + a + z e + e = f + e (Property 3) e = f (Property 2)
That is incorrect. Can you see why this is the case? Using a quick check we can see that A would never have any reason to choose the bottom row (clearly something is wrong as the question asserts it must be reduced to a 2x2 matrix!) We also notice that there we can place a straight line that separates negative and positive terms.
Your error was in your 3rd dominance argument.
The desired 2x2 matrix is:
4 -5 -2 3
This clearly cannot be reduced by using dominance arguments. The most intuitive way to see this is to think of it as an array of arrows based on what numbers are higher than others.
In this case, we can visualize there as being four arrows whereby each pair points to opposite corners. Such as array necessary for irreducibility as there are four arrows and only four possible directions with which to point. A sufficient condition would also include as array whereby the 'arrows' all point in to the tail of the proceeding one. Alternatively, you could compare 'minimax' and 'maximin' to establish 'Nash equilibrium'.
We can then go on to solve this as a linear programming problem for A by letting the proportion of the time it plays each row be p and 1-p respectively. We could also use simplex if we wanted (though, in this case, this seems rather unnecessary).
I won't write the solution up. If you want to I can check it for you?
Oh, whatever
I haven't even studied D2 yet xD
But yeah, I see where I went wrong ^.^ And yeah, I'll try and work out the rest :P
You're not differentiating correctly. dxdf(g(x))=g′(x)f′(g(x)); substitute g(x)=log(x),f(x)=exp(x) and solve for g'(x). What you have worked out is dedexp(log(x)), which is odd indeed given that e is constant.
I am not getting this at all I keep getting 1 on both sides and its driving me nuts g(x)=logxg′(x) is what we are looking for f(x)=ex f(g(x))=elogx=x f′(g(x))=1dxdf(g(x))=1
I definitely know I'm doing f′(g(x)) wrong then. Any help? Sorry for being a bit of an noob here
I am not getting this at all I keep getting 1 on both sides and its driving me nuts g(x)=logxg′(x) is what we are looking for f(x)=ex f(g(x))=elogx=x f′(g(x))=1dxdf(g(x))=1
I definitely know I'm doing f′(g(x)) wrong then. Any help? Sorry for being a bit of an noob here